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Injection molding machine clamping force calculation

by:FUHONG     2020-09-19
Clamping force is one of the important parameters of injection molding machine on the market, namely the injection molding machine on the mould clamping force. Clamping force and injection quantity, to a certain extent, reflects the machine the size of the products processing ability; And used as the main parameters of said the size of the machine specifications; Now most of the injection molding machine manufacturer with clamping force, Tons) As a machine type name ( The actual injection molding machine clamping force is greater than the standard) 。 Should try to use the model selection of the machine in maximum clamping force of the following values. Enough clamping force is proportional to the cavity of the projection area, cavity projection area is separation of projection in the mold cavity surface area. It is important to note: insufficient clamping force, can make the products in the production process of flash ( (front) Or not; And if the clamping force is too large, resulting in waste of system resources, if the mould clamping force and the template is greater than the mold needs, can make the template variant volume increase, and make the machine and the lower die life. General mould clamping force, use metric tons ( Is equal to 1000 kg) Or thousand cow dint, 1 metric ton is equal to about 10000 cows. Injection molding machine teach you clamping force effective estimation method. A, according to the vertical projection area of the injection molding products in the template, the calculation of clamping force F, clamping force, Tons) = a mould pressure ( kg/cm2) Projection area of X products ( cm2) & 划分; 1000 is F = P X. S type F = clamping force (in t) S - products in the template of vertical projection area of ( cm2) P - the pressure in the mold ( kg/cm2) 6 mm. Beg enough clamping force. A cup of ( And port) The projection area is 3. 1416×7. 9×7. 9/4 gives 49 cm = 2. The cups belong to the category of thin walled, conservative clamping force is 49 X 500 present present 1000 0. 8 = 30. 6 tons. Second, are considered in the estimation process and wall thickness is a more accurate method. Processes are dissolved material from the gate to the length of the cavity farthest point o ( See figure 1) 。 If the wall thickness of molded parts sizes, take the minimum wall thickness. For example: the same the process of hard plastic cup is 104 mm. For a more accurate clamping force. 6 = 173. 6 mm cavity pressure is 550 bar. 1000 present 0 02 x 49 members present. 8 = 34. 5 tons.
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